Mizuno reports increased excess heat

  • I'd also like to point out that multiple impolite and personalised posts .. emotional bias etc

    You are the one personalising


    When I say you have poor knowledge of heat transfer it is not personal.


    You cannot model heat transfer by radiation only. .. it is obvious.


    You have not shown any mathematical calculation that takes convection into account.

    when convection is the major heat transfer

    until you do

    your calculations based on radiation only are erroneous.

  • your calculations based on radiation only are erroneous.


    Please read and understand points 3 - 6 above.


    All I need is the (textbook) assumption that forced convection cooling is roughly proportional to temperature delta, and the (proven from insulation R value) fact that the inner insulation surface is a lot hotter than the forced air, to make my point that heat loss from enclosure is roughly proportional to radiation power.

  • I need is the (textbook) assumption that forced convection cooling is roughly proportional to temperature delta, and the (proven from insulation R value) fact that the inner insulation surface is a lot hotter than the forced air, to make my point that heat loss from enclosure is roughly proportional to radiation power.


    Your understanding of the complexity of heat transfer is poor


    which heat transfer textbook are you quoting..


    your calculations without convective heat transfer are erroneous,


    Try this for a start


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  • When I say you have poor knowledge of heat transfer it is not personal.


    You cannot model heat transfer by radiation only. .. it is obvious.


    I'm glad your comment here is not personal. Nothing in a complex system can be modelled precisely with only one component. However we gain useful insight by understanding conditions under which a given quantity is dominated by just one particular effect. I show that above.


    You will also note I'm quite careful: I use words like roughly, to first approximation, etc. In this case the extent to which forced convection can be neglected as a driver of the inner foil heat (rather than something that cools it) depends on the ration between the inner foil uplift above ambient, and the exhaust gas uplift. Roughly we have +80C for inner foil (from calculations a long time ago based on R value) and +10C for air (direct measurement). Even though the +80C figure depends on power and is very rough, my contention that radiation dominates forced convection in adding to calorimeter heat loss remains correct enough even for any inner foil uplift > 20C, supposing exit air uplift is +10C. (OK - this is a bit subtle - forced air convection cools the inner foil, but for low foil uplifts relative to forced air uplift then the cooling effect changes nonlinearly with the foil uplift. When the cooling effect is linear we can say that foil uplift will be proportional to radiation).


    PS - I define "uplift" as temperature increase relative to ambient.


  • So say that the heat is 276W/m^2, now convective cooling is around 10W/Km^3 this means that at around 30 degrees

    we can match the radiating coolant with convective cooling and this is assuming no forced convection.


    The air is around 10 degrees higher in total we have a temperature difference of 40 degrees or less we assume black body

    and low convection properties.


    I agree thar not much can pass out of the insulation with these temperatures id we assume one magnitude lower heat transfer through

    the insulation assume an area of 1m^2 we get that 30W/m^2 and a loss of 10% through the walls which should be a

    high estimate. Looks to me that Mizuno got his estemates all right.

  • Even though the +80C figure depends on power and is very rough, my contention that radiation dominates forced convection in determining calorimeter heat loss remains correct enough even for any inner foil uplift > 20C, supposing exit air uplift is +10C.

    I'm glad your comment here is not personal

    It never was.

    Your contention is erroneous


    not very rough.

    calculations a long time ago based on R value

    Please restate these calculations and show how convection has been accounted for

    remembering that the reality is that


    The convective heat transfer btw thw reactor and the calorimeter is in two parts..


    1)btw the reactor and the airstream

    2)and btw the airstream and the calorimeter wall

  • RB - would you like to say how you determine the insulation inner foil temperature? Thus, do you agree that forced air convection cools it and radiation heats it, and the balance of these two determines the temperature. And therefore to first approximation, the temperature it attains is proportional to radiation power (the heating) because forced convection cools linearly with temperature? Do I need to show you this in a text book :)


  • Stefan. Could you be a bit more precise? We are interested in what determines the heat balance at higher powers where efficiency is lower and losses are larger. There are three important temperature, all different:

    Air through calorimeter

    Reactor surface

    Insulation inner surface foil.


    Temperatures go: reactor > foil > air.


    radiation depends on reactor - foil temperature difference (and becomes highly nonlinear when this is large, so that foil temperature can be ignored)

    forced convection from foil depends on foil - air

    forced convection from reactor surface depends on reactor - air


    efficiency depends on heat loss though insulation which depends on foil temperature (maybe nonlinearly due to whole calorimeter being cooled by natural convection, but linearly according to R value of insulation if we suppose that this is more significant than the calorimeter case to air thermal resistance, I'd guess this is true at least for high powers).

  • Please restate these calculations and show how convection has been accounted for

    remembering that the reality is that


    The convective heat transfer btw thw reactor and the calorimeter is in two parts..


    1)btw the reactor and the airstream

    2)and btw the airstream and the calorimeter wall


    Glad you are understanding this. Convection is accounted by supposing:


    (1) That forced convective cooling depends on temperature difference between forced air and surface

    (2) That calorimeter inner wall - air temperature is >> air - ambient temperature.


    These two things are all that is needed for my useful statements above. When you determine an (approximate) algebraic relationship between two qtys of interest and see whether they relate to each other linearly, or as sq law, etc, it is useful in understanding systems.


    You should also note that I'm looking at the heat loss, not the heat going to the output.

  • forced convection from foil depends on foil - air

    forced convection from reactor surface depends on reactor - air


    please note that the air temperature varies significantly throughout the air stream


    from high near the reactor wall to lower near the calorimeter wall.


    and that the convective HTC varies greatly with airspeed


    which is variable (0-2 m/s) throughout the entire box.


    which makes any calculation with convective HTCs difficult


    and therefore any calculation of the temperatures based on textbook generalised assumptions

    are extremely prone to assumption error

    THHnew..

    if one reads a standard engineering textbook on heat transfer

    it will state that the actual HT , both radiative and convective

    depends greatly on the details of geometry.


    In addition your calculations of box temperature based on Stefan's Law are

    erroneous. If you do not take into account the reradiated radiation from the

    calorimeter boxwalls you get a calorimeter box temperature of 80C..


    if you do take into account the reradiated radiation you get a significantly lower

    box wall temperature

  • In addition your calculations of box temperature based on Stefan's Law are erroneous

    For instance if the reactor/box configuration is idealized as concentric cylinders

    the power output assuming the ambient is 80C is 352W

    but when the reflected heat from the box is accounted for the

    net power output is only 215 W.


    The actual net radiant power from the reactor when the shape factors are included is something like 150W.


  • RB: I agree with some of your speculation and generalisation above.


    However: what evidence do you have that air temperature is lower near the foil? The boundary layer by the foil will be heated by the foil and thus high in temperature than the main air body.


    Otherwise you are saying that HTCs here are difficult to calculate precisely. I agree.


    Luckily my argument does not depend on that. Radiation from reactor to foil is relatively insensitive to geometry, depending only on the "view" surface area, the emissivities, and the temperatures, of these two components. Much easier than convective HTC. If you have an example of how in this case the radiation depends much on the geometry - rather than the overall "view" surface area I'd be interested. It is a neat fact that for radiation all that matters is how much of the "other" surface you can see, not what angle it is at (this cancels if you take view area rather than surface area) nor the separation.


    You are wrong about the re-radiation. That is because at the temperatures I did this calculation (380C reactor vs 80C wall) the re-radiation is less that 10% of the radiation due to the T^4 factor for relatively small gaps. If you get a small reactor inside a large box that increases the foil to reactor / reactor to foil ratio by the difference in the view factors but we still have radiation from reactor to foil dominating.


    (270+80)^4 ~ 15E9 vs

    (270+380)^4 = 178E9


    However it is true that this re-radiation factor is one thing that makes the radiation component highly nonlinear (even worse than the T^4). I commented on this in my original post on the subject: because it shows that assuming linearity of all things in this system is highly dangerous. It is easy to neglect radiation because at low temperature differences these factors usually combine to make it much smaller than convection. Of course in this case, when we are interested in how much the insulation heats up, radiation is significant even at quite low temperatures (for example if radiating surface is +100C from ambient, the air flow is at most +5C from ambient, probably quite a bit less, and even taking into account more significant re-radiation here, I'd ballpark guess the foil temperature is more determined by radiation than convection).




    THH

  • PS - the box inner foil temperature of ~ 80C is experimentally determined, if you believe the efficiency and heat output data in the paper, since the insulation R value is known. The only unknown is the amount the whole calorimeter housing will heat up - but that affect would make the foil higher in temperature if significant.


    I do wonder about temperature errors due to conduction and radiation onto the heat sensor from the whole housing - if it does heat up.

  • That is because at the temperatures I did this calculation (380C reactor vs 80C wall) the re-radiation is less that 10% of the radiation due to the T^4 factor for relatively small gaps.

    This is wrong.

    Using the accepted method..

    The effect of the foil is to reduce the output power for 380C reactor/80C wall situation

    from 377 W to 228W.


    This difference(377-228) is 40%.... not 10%.

    I guess this is why people use aluminum foil for insulation.


    Your calculations of 80C wall temperature are erroneous.

    I get 40-41 C.

  • Assume that the convective transfer is the same inside as on the outside.


    then heat from box to outside air is


    Q1 = K (T1 - T0) A


    Heat to inside air is (air is 10 degrees hotter)


    Q2 = K (T2 - 10 - T0) A


    in balance the heat flowing through the insulation is tha same as Q1 e.g


    Q1 = K2 (T2 - T1) A


    Assume K2 = 0.1K


    Q1 = 0.1 K (T2 - T1) A = K(T1 - T0)A


    or

    0.1 (T2 - T1) = T1 - T0


    or

    1.1T1 = 0.1 T2 + T0


    or aprox


    T1 = 0.1T2 + T0 - 0.1 T0


    T0 = 20, T2 = 80 => T1 = 30


    Hence

    Q1 = K*(30-20)*A = K 10 A = 0.1 K (T2 - T0) A


    Now A = 2m^2 we get with K=10

    Q1 = 200W


    Q2 = K *(80-10-20) A = 10 * 50 * 2 = 1000W


    Contradiction, K=5 => Q2 = 500W, contradiction as well.


    In fact


    Q1/Q2 = 0.1 (T2 - 20) / (T2 - 30) ~ 0.1-0.2


    So The heat from the wall to the inside air is about 10x the heat to the outside air.


    150W goes to the wall, that means that most of it goes to the inside air so Q2 ~ 150W


    This means that Q1 ~ 15W - 30W


    Also,


    Q2 = K (T2 - 30) * 2 = 150


    K = 10 => T2 ~ 38 degrees

    K = 5 => T2 ~ 45 degrees


    /Stefan

  • This is my model with assumptions.

    box wall temp =41.

    Airflow splits into 1/10 to around the reactor, 9/10 around the box.

    average air temp around reactor = 80c.... box air flow ave temp = 30.

    versus box T=41C reactor T=380C...

    these two flows unite to give 35C outlet.


    losses from box to ambient

    radiative = 0.03 x2x sigma x{ 41/23.,K,4th delta} = 7W

    convective = 2x2x 41/30 delta =72W


    input =50+250w from reactor.

    radiative HT from reactor to box =0.1215x 0.2x{ 380/41.,K,4th delta} x 0.7 shape factor =166W

    (effective emissivity =0.1215... includes the effect of foil reflected heat)

    convective HT from reactor to air = 2.2x 0.2 x 380/80 delta= 134W


    convective HT from box to air = 4x2x 41/30 delta =88W


    measured HT from blower outlet = air mass flowrate x cp x 35/23 deltaT =222W


    Check 1. 300 =222+7+72? =299. Yes OK

    Check 2 . Convective HTs = blower HT?

    88+134 =222W Yes OK.


    Some problems with HTCs for forced convection 4 versus 2.2

    also assumption of the shape factor=0.7..( small reactor cylinder inside big box with reactor bottom side pointing at brick insulation)

    Lots of assumptions.. averaged tempertures.. plenty of errors are possible



    Higher wall temps give excessive losses from box to ambient

    example ... 80c gives

    radiant loss= 27W

    convective loss = 228W

    these two together produce an xs loss of 183W to ambient... with the given input of 300W and the 222W blower outlet output.

    Lower temps give an excess of input heat over output heat..eg 39C gives an excess of 8W


    Average Box wall T of 41 seems OK given the stated assumptions.

  • Wow RB! That is interesting. I'd like to check your figures though. That is because if you are right there is a serious hole in Mizuno's data for calorimeter efficiency, or Jed's data for the R value of the insulation, or mizuno's data for how the reactor temperature depends on the power. And I agree that I was ignoring the (large) effect of e2 on the transferred power. I was thinking of the re-radiation, but ignoring the original reflected radiation component, of the foil! If it has v low emissivity it reflects nearly all of the incident radiation.


    An insulated acrylic box is used for airflow calorimetry. It is 400 mm × 750 mm, height 700 mm. During a test, the
    inside of the plastic box is covered with 1.91 m2 of reflective padded aluminum insulation (shanetsu.com, Fig. 7).
    This minimizes losses to radiation. These losses are low in any case, because the cooling air keeps the inside of the
    box at ∼36◦C (16◦C above ambient). Similar insulation from a US vendor (US Energy Products) has an R-value of 11,
    so the insulation radiates ∼3 W (16◦C/11 W/m2 × 1.9 m2


    Jed has confirmed this R value is imperial (and hence the calculation here wrong): to get metric r value we divide by 5.67:

    R = 1.94 C/Wm^2


    To get the inner temperature of the insulation walls we note from above that the total insulated area is 1.91m^2, and therefore the thermal characteristics of the box are:


    deltaT = 1.01C/W


    Figure 10 shows the reactor temperature vs. the heat recovery rate. When there is no input power, and the reactor
    body temperature is 25◦C, the recovery rate should be close to 1. When the reactor body temperature is 100◦C, the
    recovery rate is 0.93; it is 0.82 at 300◦C, and 0.78 at 360◦C. Naturally this is the same in the test reactor as the
    calibration reactor; the recovery rate decreases as the temperature of the test reactor rises.


    From Table 1:

    4/11

    3199.0

    Pin = 200.9 W

    Pexcess = 96.74W

    Treactor = 383.93C


    We have total output of 298W, and reactor temperature of 383C, at which efficiency is 78% (a little less given shape of graph).


    Thus total heat loss from calorimeter here is 65W.


    With the corresponding deltaT, and outside of calorimeter = 30C (it might well be hotter) we have a temperature of 95C on the inside of the calorimeter insulation (the foil).


    I agree with RB that my figures for radiation were too high - unless the inner foil has become much higher emissivity (e.g. because of scorching etc).


    Let us suppose that he is correct about radiation (the theoretical model used is not right, because the reactor is not in the middle of the calorimeter, but it is not too bad).


    Convection cannot possibly heat the reactor foil above the output air temperature, which is given as a 2.8C rise for 50W, or 17C for 300W. That is assuming the 300W figure from the paper is NOT adjusted for efficiency. I think it is adjusted, in which case we have approx 234W detected at output or 13C rise.


    We thus have the 65C rise coming from radiation, but not nearly enough radiation to do this.


    Anyone got a solution?


    possibles:


    The insulation is much less good than the claimed R=11 (metric), allowing a (say) 36C internal wall temperature, as Mizuno suggests in the paper, to drive this 65W heat loss

    The heat loss comes from somewhere other than the box sides

    The efficiency calibration is wrong

    The heat output is wrong

    The radiation is in fact a lot higher due to reactor closer to walls, or foil having very high emissivity, or reactor having very high emissivity.


    THH

  • THHuxley and Robert Bryant,


    You're both word fighting/arguing over how much excess heat R20 makes when you both agree that R20 makes the output air hotter by 13C vs 3C for the control on the same 50 Watts input from the presented data in figure 5.


    Irregardless, it is an astounding result that will likely soon be replicated by multiple parties. Hotter MUST be excess power from the reaction (assuming the thermal characteristics of the control vs. the active reactor in the calorimeter are substantially similar). Sustained for sufficient time, it must be non-chemical excess energy. That is a revolutionary result. Only excess power makes things HOTTER if the other thermal characteristics are the same. It must be non-chemical, i.e. LENR.


    A better calorimeter can and will be built later by other scientists and engineers looking to exactly quantify the LENR effect. Arguing over the first calorimeter's accuracy/inaccuracy when the results is clearly HOTTER is like not seeing the forest for the weeds.


    If you continue investing your time in this technical disagreement, can you both please tone it down and make it less personal. (I.e. eliminate personal and personal possessive pronouns when critiquing the other's work.) Thank you.

  • You're both word fighting/arguing over how much excess heat R20 makes


    I am not..

    I have already said to replicate R20 because it shows 300% delta T extra for active reactor versus calibration

    I dispute that the insulation is 80C . I get 41C

    Apparently THHnew has now adjusted his position because he has used an erroneous method

    for calculating Stefan' s Law effects


    can you both please tone it down and make it less persona


    It is not personal.in fact it is rather boring.. it is about the correct method for calculation of Q12.

    The areas and emissivities of both bodies need to be taken into account

    This method corrects for the effect of the reflected heat from the foil, which is not 10% but about 40%



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